Mark Scheme (Results) January 013 GCE Core Mathematics C (6664/01)
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General Marking Guidance All candidates must receive the same treatment. Eaminers must mark the first candidate in eactly the same way as they mark the last. Mark schemes should be applied positively. Candidates must be rewarded for what they have shown they can do rather than penalised for omissions. Eaminers should mark according to the mark scheme not according to their perception of where the grade boundaries may lie. There is no ceiling on achievement. All marks on the mark scheme should be used appropriately. All the marks on the mark scheme are designed to be awarded. Eaminers should always award full marks if deserved, i.e. if the answer matches the mark scheme. Eaminers should also be prepared to award zero marks if the candidate s response is not worthy of credit according to the mark scheme. Where some judgement is required, mark schemes will provide the principles by which marks will be awarded and eemplification may be limited. When eaminers are in doubt regarding the application of the mark scheme to a candidate s response, the team leader must be consulted. Crossed out work should be marked UNLESS the candidate has replaced it with an alternative response. Unless indicated in the mark scheme a correct answer with no working should gain full marks for that part of the question.
General Instructions for Marking EDEXCEL GCE MATHEMATICS 1. The total number of marks for the paper is 75.. The Edecel Mathematics mark schemes use the following types of marks: M marks: method marks are awarded for knowing a method and attempting to apply it, unless otherwise indicated. A marks: Accuracy marks can only be awarded if the relevant method (M) marks have been earned. B marks are unconditional accuracy marks (independent of M marks) Marks should not be subdivided. In some instances, the mark distributions (e.g., B1 and ) printed on the candidate s response may differ from the final mark scheme. 3. Abbreviations These are some of the traditional marking abbreviations that will appear in the mark schemes and can be used. bod benefit of doubt ft follow through the symbol will be used for correct ft cao correct answer only cso - correct solution only. There must be no errors in this part of the question to obtain this mark isw ignore subsequent working awrt answers which round to SC: special case oe or equivalent (and appropriate) dep dependent indep independent dp decimal places sf significant figures The answer is printed on the paper The second mark is dependent on gaining the first mark 4. All A marks are correct answer only (cao.), unless shown, for eample, as ft to indicate that previous wrong working is to be followed through. After a misread however, the subsequent A marks affected are treated as A ft, but incorrect answers should never be awarded A marks. 5. For misreading which does not alter the character of a question or materially simplify it, deduct two from any A or B marks gained, in that part of the question affected. 6. If a candidate makes more than one attempt at any question:
If all but one attempt is crossed out, mark the attempt which is NOT crossed out. If either all attempts are crossed out or none are crossed out, mark all the attempts and score the highest single attempt. 7. Ignore wrong working or incorrect statements following a correct answer. 8. The maimum mark allocation for each question/part question(item) is set out in the marking grid and you should allocate a score of 0 or 1 for each mark, or trait, as shown: am aa b b bb bm ba 0 1
January 013 6664 Core Mathematics C Mark Scheme Question Number Scheme 1. ( 5) 6 6 ( ) = 64 Award this when first seen (not 64 0 ) B1 6 5 + 6 ( ) ( 5 ) + ( ) ( 5 ) 5 4 Attempt binomial epansion with correct structure for at least one of these terms. E.g. a 6 p ( 5 ) p term of the form: ( ) 6 with p = 1 or p = consistently. Condone sign errors. Condone missing brackets if later work implies correct structure and allow alternative forms for binomial coefficients 6 6 p Marks e.g. 6 C or or even 1 1 1 960 Not + 960 (first) ( + )6000 (Second) (4) Way 64( 1 ±...) 64 and (1 ±... Award when first seen. B1 6 5 5 6 5 5 1 = 1 6 + Correct structure for at least one of the underlined terms. E.g. a term of the form: 6 ( k) p p p p consistently and k ± 5 with = 1or = Condone sign errors. Condoned missing brackets if later work implies correct structure but it must be an epansion of ( 1 k) 6 where k ± 5 960 Not + 960 ( + )6000 (4)
Question Number Scheme Marks. (a) f(1) = a+ b 4 3= 0 or a + b 7 = 0 Attempt f(±1) a + b = 7 * Must be f(1) and = 0 needs to be seen 3 (b) a( ) b( ) ( ) f( ) = + 4 3= 9 Attempt f(±) and uses f(±) = 9 8a+ 4b+ 8 3= 9 ( 8a + 4b = 4) Solves the given equation from part (a) and their equation in a and b from part (b) as far as a =... or b =... Correct equation with eponents of (-) removed a = and b = 5 Both correct () (a) (b) Long Division 3 a + b 4 3 1 = a + p + q ( ) ( ) where p and q arein terms of a or b or both and sets their remainder = 0 ( ) ( ) NB Quotient = 4 a + a + b + a + b a + b = 7 * ( 4 3) ( ) 3 a b a p q + + = + + where p and q arein terms of a or bor both and sets their remainder = 9 ( ) ( ) NB Quotient = 4 4 a + b a + a b 4b 8a + 5 = 9 Follow scheme for final marks (4) [6] ()
3. (a) (b) 3 10000 (1.05) = 138915 * n 1 10000 (1.05) 00000 Or 10000 1.05 1.05 1.05 = 138915 Or 10000, 16000, 13300, 138915 Or a = 10000 and a ( 1.05) 3 = 138915 > Allow n or n 1 and >, <, or = etc. n 1 5 log1.05 > log 3 5 log ( ) 3 n 1 > or equivalent e.g ( n > ) log1.05 7 log 4 log1.05 04 Takes logs correctly Allow n or n 1 and >, <, or = etc. Allow n or n 1 and >, <, or = etc. Allow 1.6 & or awrt 1.67 for 5/3. : Identifies a calendar year using their value of n or n -1 : 04 B1 (1) (c) 11 ( ) n a(1 r ) 10000 1 1.05 = 1 r 1 1.05 : Correct sum formula with n = 10, 11 or 1 : Correct numerical epression with n = 11 1704814 Cao (Allow 1704814.00) Listing or trial/improvement in (b) U 10 = 186 159.39, U 11 = 195 467.36, U 1 = 05 40.7 Attempt to find at least the 10 th or 11 th or 1 th terms correctly using a common ratio of 1.05 (all the terms need not be listed) Forms the geometric progression correctly to reach a term > 00 000 Obtains an 11 th term of awrt 195 500 and a 1 th term of awrt 05 00 Uses their number of terms to identify a calendar year 04 (5) (3) [9] (5)
4. ( ) 1 cos 0.4 = 113.58 ( α) Awrt 114 B1 Uses their α to find. α + 10 3 10= α = α ± 10 α 3 Allow = not ± 10 3 3 = 41. Awrt ( ) 3 10 = 360 α (46.4...) 360 α (can be implied by 46.4...) = 85.5 Awrt 3 10 = 360 + α = 473.57... 360 + α (Can be implied by 473.57...) ( ) ( ) = 161. Awrt
5. (a) (i) The centre is at (10, 1) B1: = 10 B1: y = 1 B1 B1 (ii) (a) Way Uses ( 10) + ( y 1) = 195 + 100 + 144 r =... Completes the square for both and y in an attempt to find r. ( ± "10") ± aand ( y± "1") ± band + 195 = 0, a, b 0 ( ) Allow errors in obtaining their r but must find square root r = 10 + 1 195 A correct numerical epression for r including the square root and can implied by a correct value for r r = ± r = 7 Not 7 unless 7is rejected Compares the given equation with + y + g + fy + c = 0 to write down centre ( g, f) i.e. (10, 1) Uses r ( "10") ( "1") B1: = 10 B1: y = 1 B1B1 = ± + ± c r = 10 + 1 195 A correct numerical epression for r r = 7 (5) (5) (b) MN = (5 "10") + (3 "1") Correct use of Pythagoras MN ( = 65) = 5 (c) NP = ("5" "7" ) NP MN r = ( ) () (c) Way NP ( = 576 ) = 4 7 cos ( NMP) = NP = "5"sin ( NMP) "5" Correct strategy for finding NP NP = 4 () () [9]
6. (a) log( 15) log( 15) + = + B1 ( 15) log( 15) log log + + = Correct use of log a log b = log a b 6 = 64or log 64 = 6 64 used in the correct contet B1 ( + 15) ( + 15) log = 6 = 64 Removes logs correctly + + = 30 5 64 or + + = 1 30 5 64 Must see epansion of ( + 15) to score the final mark. 34+ 5 = 0 * (b) ( 5)( 9) = 0 = 5 or = 9 : Correct attempt to solve the given quadratic as far as =... : Both 5 and 9 (5) () [7]
7. (a) 9 = 4 + 6 4 6cosα cos α =... XY 4 + 6 9 9 (a) Way Correct use of cosine rule leading to a value for cos α cosα = = = 0.604.. 4 6 48 α =. Cso (. must be seen here) (NB α =.19516005) () Correct use of cosine rule = 4 + 6 4 6cos. XY =.. leading to a value for XY XY = 81.01... XY = 9.00... (b) π.( = 4.06366...) π. or awrt 4.06 B1 1 Correct method for major 4 "4.06" sector area. 3.5 Awrt 3.5 (b) Way (c) Circle Minor sector π 4 Correct epression for circle area B1 1 Correct method for π 4 4. = 3.5 circle - minor sector area = 3.5 Awrt 3.5 Area of triangle = Correct epression for the area of 1 4 6 sin.( = 9.56) triangle XYZ B1 So area required = 9.56 + 3.5 Their Triangle XYZ + (part (b) answer or correct attempt at major sector) = 4.1 cm or 4.0 cm Awrt 4.1 or 4.0 (Or just 4) () (3) (3) (3) (d) Arc length = 4 4.06( = 16.4) Or 8π 4. : 4 their ( π.) Or circumference minor arc ft : Correct ft epression Perimeter = ZY + WY + Arc Length 9 + + Any Arc Perimeter = 7. or 7.3 Awrt 7. or awrt 7.3 (4) [1]
4 8. y = 6 3 3 (a) dy 1 = 3+ or 3+ 1 4 d 4 dy 1 = 0 3 + = 0 =... or 4 d dy 1 3 4 d = + : n n 1 1 0 3 4 ( or or6 0) : Correct derivative y = 0 and attempt to solve for May be implied by dy 1 1 = 3+ = 0 = 3 =... or 4 4 d Substitutes = into their y So 4 = 4and= or ( ) ( ) 4 dy 1 = 3+ or 3+ 1 = 0 4 d Correct completion to answer with no errors by solving their y = 0 or substituting = into their y (b) = Awrt -1.41 B1 (c) (d) d y 48 = 5 or 48 d 5 Follow through their first derivative from part (a) An appreciation that either y > 0 a minimum B1 or y < 0 a maimum Maimum at P as y < 0 Cso B1 Need a fully correct solution for this mark. y need not be evaluated but must be correct and there must be reference to P or to and negative or < 0 and maimum. There must be no incorrect or contradictory statements (NB allow y = awrt-8 or -9) Minimum at Q as y > 0 Cso B1 Need a fully correct solution for this mark. y need not be evaluated but must be correct and part (b) must be correct and there must be reference to P or to and positive or > 0 and minimum. There must be no incorrect or contradictory statements (NB allow y = awrt 8 or 9) Other methods for identifying the nature of the turning points are acceptable. The first B1 is for finding values of y or dy/d either side of or their at Q and the second and third B1 s for fully correct solutions to identify the maimum/minimum. B1ft (4) (1) (1) (3) [9]
16 9. y = 7 9 (a) 6.7, 3.634 B1, B1 (b) 1 1 1 or 4 B1 Need {} or implied...{(0 + 0) + ( 5.866 + "6.7" + 5.10 + "3.634" + 1.856) } later for ft ft () 1 0.5 (0 0) 5.866 "6.7" 5.10 "3.634" 1.856 + + + + + + { ( )} = 1 45.676 4 = 11.4 cao (c) yd= 3 ( ) 7 6 + 16 1 + c 3 1 ( 7( 4) ( 4) 6( 4) + 16( 4) ) 3 1 ( 7() 1 () 1 6() 1 + 16() 1 ) n n 1 : + on any term : 7 3 : 6 : + 16 1 Attempt to subtract either way round using the limits 4 and 1. Dependent on the previous (4) d = (48 36) 1 Cao (6) [1]
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